8
$\begingroup$

In some literature on linear algebra determinants play a critical role and are emphasized in the earlier chapters (see books by Anton & Rorres, and Lay). However in other literature it is totally ignored until the latter chapters (see Gilbert Strang). How much importance should we give the topic of determinants? I tend to use it to find linear independence of vectors and might extend this to finding the inverse but I think Gauss Jordan and LU might be easier for inverse. Does it have any other uses in Linear Algebra. Are there areas where determinants are used and have a real impact? Are there any real life applications of determinants? Is there a really good motivating example or explanation which will hook students into this topic? In linear algebra, where should determinants be placed? Like I said in my comment - in some literature it is at the beginning whilst in others it is bolted on at the end. I like the idea of checking if vectors are independent by using determinants so think they should be placed before independence of vectors. What do you think? If you teach a linear algebra course where do you place this topic.

$\endgroup$
3
  • 1
    $\begingroup$ Define "real" in "real impact"...Determinants appear in many places in algebra (not only linear but abstract), in analysis (Hessian, Jacobian, Wronskian, etc.), in differential equations, etc., etc., etc..... $\endgroup$
    – DonAntonio
    Jun 20, 2012 at 13:59
  • $\begingroup$ As Peter pointed out, from the computational point of view, a determinant is expensive, so people try to avoid to compute it when dealing with large matrices. On the theoretical point of view, though, it plays an important role in many fields. $\endgroup$
    – bartgol
    Jun 20, 2012 at 14:09
  • $\begingroup$ Before the eigenvalues-eigenvectors chapter. You need determinants to write the characteristic polynomial of a matrix... $\endgroup$
    – Bob
    Jul 29, 2018 at 15:45

6 Answers 6

12
$\begingroup$

You are witnessing a shift in emphasis away from determinants. This does not mean they are unimportant; on the contrary, they are quite important – think of the change of variable formula in multiple integrals, for instance – but by introducing them too early in linear algebra courses, and spending too much time on their properties, we have encouraged students to use determinants where their use is not appropriate, such as in solving linear systems. You will find the most extreme cases in old linear algebra texts, introducing determinants by their closed formula, which requires on the order of $n!$ operations to compute an $n\times n$ determinant.

$\endgroup$
6
$\begingroup$

There are several ways to motivate the determinant but I can think of no more interesting approach than the essential feature which characterizes it: Every multilinear alternating form defined on the product $(R^n)^n := R^n \times \cdots \times R^n$ is a scalar multiple of the determinant function. In fact, the determinant function is the unique multilinear alternating function when evaluated on the Identity matrix, treating columns as vectors, yields unity. From this characterization, one can formally derive the admittedly horrendous-looking combinatorial formula. For illustration purposes, for lower dimensions, say n=2 or n =3, it is tractable to compute the determinant explicitly using this characterization. This is preferable in my opinion than stating the formula and showing that said formula satisfies properties A-Z.

The determinant shows up in a surprising number of places and for this reason alone is important. I think though it might be an excellent place to introduce the idea that mathematical objects are not necessarily important for what they are but, rather, the properties they satisfy (e.g., universal properties). If you think linear algebra is a good place to introduce the basic concepts of abstract mathematics I think the determinant does this quite nicely. The description of the determinant above is elementary enough that anyone taking linear algebra should be able to understand it and proving subsequent theorems about determinants goes significantly smoother with this characterization as opposed to trying to pound it out with the combinatorial formula.

$\endgroup$
1
  • $\begingroup$ Thanks for the detailed answer. $\endgroup$
    – matqkks
    Jun 21, 2012 at 9:21
5
$\begingroup$

This is quite an informal answer.

Determinants basically help to describe the nature of solutions of linear equations. The determinant of a real matrix is just some real number, telling you about the invertibility of the matrix and hence telling you things about linear equations wrapped up in the matrix.

The determinant being non-zero is equivalent to the matrix being invertible, which is equivalent to the corresponding sets of linear equations having EXACTLY one solution.

The determinant being zero means the matrix is NOT invertible. In this case the corresponding sets of linear equations can either have infinitely many solutions or none at all (depending on the numbers on the RHS).

So really the determinant is useful anywhere that linear equations crop up. For example, when checking linear independence, this is the same as demanding the existence of a UNIQUE solution to a set of linear equations (i.e. the zero vector solution). This is the same as the matrix determinant being non-zero as discussed above. Linear dependence must therefore be the same as the determinant being zero (so that there may be non-zero solutions to the equations, i.e. so that some of the vectors really can be made to add up non-trivially to give one of the others).

$\endgroup$
1
  • $\begingroup$ Yes, but the OP is just asking for justification that the determinant is a useful creation. I could have given many other examples where determinants are "used" but really you come across them as you learn more. Yes there are better/worse ways to do specific things but this wasn't what was asked... $\endgroup$
    – fretty
    Jun 22, 2012 at 22:06
0
$\begingroup$

Funny that nobody mentioned that the determinant of a nxn matrix with real entries is the signed hypervolume of the parallelotope generated by the n columns of the matrix. This means that det = 0 (i.e. hypervolume = 0) if and only if the columns of the matrix are linearly dependent. It may be useful to meditate on the relationship between this interpretation of a determinant and the definition given earlier (the only n-linear alternating function of the columns of the matrix which equals 1 when evaluated on the identity matrix) and also on why a determinant appears in the change of variable formula for multiple integrals.

$\endgroup$
0
$\begingroup$

The determinant is important because, among other things, leads to the characteristic polynomial of a matrix.

Indeed—notice that $Ax = \lambda x$, $x \ne 0$ if and only if $p(\lambda) := \det(\lambda I_n - A) = 0$.

$\endgroup$
0
$\begingroup$

Determinants do have applications in the real world as the encode certain geometric properties of the transformation. Firstly, they have a way of encoding information about how volumes change with respect to the transformation. This is easy to see from the multilinearity and for simplicity we'll just use $\mathbb{R}^3$ with the canonical basis vectors which define a $1\times 1 \times 1$ cube. If I scale any coordinates, say the first, so that we get a $2 \times 1 \times 1$ rectangular prism and the volume has doubled. In fact, I can scale the first coordinate by anything and it's equivalent to multiplying the volume by the number. The same is true for the second and third coordinate as well.

This means the volume is scales linearly in each coordinate but we don't have a way to interpret negative numbers as negative volumes don't make a lot of sense. However we don't have to interpret the negatives as negative volumes, merely the usual volume but with a reversal in orientation. From here you're not far from the determinant as an alternating multilinear map is a natural choice. It's not obvious that full multilinearity work for the volume when considering shear transformations but it does do the trick.

Mathematically however, the geometric properties are less interesting than the algebraic ones. In particular I like that determinants give you group homomorphism $\det: GL_n(\mathbb{F}) \rightarrow \mathbb{F}$ and I also think it's really cool that it maps all the zero divisors in the ring of $n \times n$ matrices to $0$ as well. It's sort of collapses the vector space back down into the field it came from in a neat little package.

Arguably the most important role determinants play is with regard to the characteristic polynomial which arises when calculating eigenvectors and eigenvalues. The Cayley-Hamilton theorem is a major result along these lines which makes determinants particularly important in this regard.

As for where I'd place determinants, I prefer Strang's method of deferring them. They're best understood as an alternating multilinear form and those properties give you all the tools you need to calculate them as long as you understand linearity. The tensor product will also be a multilinear so the determinate provides a simple introduction to that basic idea. Between this and the characteristic polynomial I feel it's too important to skip determinants, even for students who are going on to engineering or the sciences. There is more here than just a plug-and-chug formula to calculate volumes, but a important mathematical object in its own right.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .